Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $a = \dfrac{q^2 - 5q - 14}{-q + 7} \times \dfrac{2q + 6}{-8q - 16} $
First factor the quadratic. $a = \dfrac{(q + 2)(q - 7)}{-q + 7} \times \dfrac{2q + 6}{-8q - 16} $ Then factor out any other terms. $a = \dfrac{(q + 2)(q - 7)}{-(q - 7)} \times \dfrac{2(q + 3)}{-8(q + 2)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ (q + 2)(q - 7) \times 2(q + 3) } { -(q - 7) \times -8(q + 2) } $ $a = \dfrac{ 2(q + 2)(q - 7)(q + 3)}{ 8(q - 7)(q + 2)} $ Notice that $(q - 7)$ and $(q + 2)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ 2\cancel{(q + 2)}(q - 7)(q + 3)}{ 8(q - 7)\cancel{(q + 2)}} $ We are dividing by $q + 2$ , so $q + 2 \neq 0$ Therefore, $q \neq -2$ $a = \dfrac{ 2\cancel{(q + 2)}\cancel{(q - 7)}(q + 3)}{ 8\cancel{(q - 7)}\cancel{(q + 2)}} $ We are dividing by $q - 7$ , so $q - 7 \neq 0$ Therefore, $q \neq 7$ $a = \dfrac{2(q + 3)}{8} $ $a = \dfrac{q + 3}{4} ; \space q \neq -2 ; \space q \neq 7 $